Question

Answer 1

Có: \(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)

\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\left(đpcm\right)\)

Answer 2

1/6 + 1/12 + 1/20 + ... + 1/9900`

`= 1/(2.3) + 1/(3.4) + 1/(4.5) + ... 1/(99.100)`

`= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100`

`= 1/2 - 1/100 < 1/2`